Practicing Success
The value of integral $\int\limits^{1}_{0}\sqrt{1-x^2}dx$ is : |
$\frac{\pi}{2}$ $\frac{\pi}{3}$ $\frac{\pi}{4}$ $\pi $ |
$\frac{\pi}{4}$ |
The correct answer is Option (3) → $\frac{\pi}{4}$ $I=\int\limits^{1}_{0}\sqrt{1-x^2}dx$ let $x=\sin θ$ $dx=\cos θdθ$ $x → 0, θ → 0$ $x → 1, θ → \frac{π}{2}$ $I=\int\limits^{\frac{π}{2}}_{0}\sqrt{1-\sin^2θ}\cos θdθ$ $=\int\limits^{\frac{π}{2}}_{0}\cos^2θdθ=\int\limits^{\frac{π}{2}}_{0}\frac{\cos 2θ}{2}+\frac{1}{2}dθ$ $=\left[\frac{\sin 2θ}{4}+\frac{θ}{2}\right]^{\frac{π}{2}}_{0}=\frac{\pi}{4}$ |